# Wallis' integrals

In mathematics, and more precisely in analysis, the Wallis integrals constitute a family of integrals introduced by John Wallis.

## Definition, basic properties

The Wallis integrals are the terms of the sequence ${\displaystyle (W_{n})_{n\geq 0}}$ defined by

${\displaystyle W_{n}=\int _{0}^{\frac {\pi }{2}}\sin ^{n}x\,dx,}$

or equivalently (by the substitution ${\displaystyle x={\tfrac {\pi }{2}}-t}$),

${\displaystyle W_{n}=\int _{0}^{\frac {\pi }{2}}\cos ^{n}x\,dx.}$

The first few terms of this sequence are:

 ${\displaystyle W_{0}}$ ${\displaystyle W_{1}}$ ${\displaystyle W_{2}}$ ${\displaystyle W_{3}}$ ${\displaystyle W_{4}}$ ${\displaystyle W_{5}}$ ${\displaystyle W_{6}}$ ${\displaystyle W_{7}}$ ${\displaystyle W_{8}}$ ... ${\displaystyle W_{n}}$ ${\displaystyle {\frac {\pi }{2}}}$ ${\displaystyle 1}$ ${\displaystyle {\frac {\pi }{4}}}$ ${\displaystyle {\frac {2}{3}}}$ ${\displaystyle {\frac {3\pi }{16}}}$ ${\displaystyle {\frac {8}{15}}}$ ${\displaystyle {\frac {5\pi }{32}}}$ ${\displaystyle {\frac {16}{35}}}$ ${\displaystyle {\frac {35\pi }{256}}}$ ... ${\displaystyle {\frac {n-1}{n}}W_{n-2}}$

The sequence ${\displaystyle (W_{n})}$ is decreasing and has positive terms. In fact, for all ${\displaystyle n\geq 0:}$

• ${\displaystyle W_{n}>0,}$ because it is an integral of a non-negative continuous function which is not identically zero;
• ${\displaystyle W_{n}-W_{n+1}=\int _{0}^{\frac {\pi }{2}}\sin ^{n}x\,dx-\int _{0}^{\frac {\pi }{2}}\sin ^{n+1}x\,dx=\int _{0}^{\frac {\pi }{2}}(\sin ^{n}x)(1-\sin x)\,dx>0,}$ again because the last integral is of a non-negative continuous function.

Since the sequence ${\displaystyle (W_{n})}$ is decreasing and bounded below by 0, it converges to a non-negative limit. Indeed, the limit is zero (see below).

## Recurrence relation

By means of integration by parts, a recurrence relation can be obtained. Using the identity ${\displaystyle \sin ^{2}x=1-\cos ^{2}x}$, we have for all ${\displaystyle n\geq 2}$,

{\displaystyle {\begin{aligned}\int _{0}^{\frac {\pi }{2}}\sin ^{n}x\,dx&=\int _{0}^{\frac {\pi }{2}}(\sin ^{n-2}x)(1-\cos ^{2}x)\,dx\\&=\int _{0}^{\frac {\pi }{2}}\sin ^{n-2}x\,dx-\int _{0}^{\frac {\pi }{2}}\sin ^{n-2}x\cos ^{2}x\,dx.\qquad {\text{Equation (1)}}\end{aligned}}}

Integrating the second integral by parts, with:

• ${\displaystyle u'(x)=\cos(x)\sin ^{n-2}(x)}$, whose anti-derivative is ${\displaystyle u(x)={\frac {1}{n-1}}\sin ^{n-1}(x)}$
• ${\displaystyle v(x)=\cos(x)}$, whose derivative is ${\displaystyle v'(x)=-\sin(x)}$

we have:

${\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{n-2}x\cos ^{2}x\,dx=\left[{\frac {\sin ^{n-1}x}{n-1}}\cos x\right]_{0}^{\frac {\pi }{2}}+{\frac {1}{n-1}}\int _{0}^{\frac {\pi }{2}}\sin ^{n-1}x\sin x\,dx=0+{\frac {1}{n-1}}W_{n}.}$

Substituting this result into equation (1) gives

${\displaystyle W_{n}=W_{n-2}-{\frac {1}{n-1}}W_{n},}$

and thus

${\displaystyle W_{n}={\frac {n-1}{n}}W_{n-2},}$

for all ${\displaystyle n\geq 2.}$

This is a recurrence relation giving ${\displaystyle W_{n}}$ in terms of ${\displaystyle W_{n-2}}$. This, together with the values of ${\displaystyle W_{0}}$ and ${\displaystyle W_{1},}$ give us two sets of formulae for the terms in the sequence ${\displaystyle (W_{n})}$, depending on whether ${\displaystyle n}$ is odd or even:

• ${\displaystyle W_{2p}={\frac {2p-1}{2p}}\cdot {\frac {2p-3}{2p-2}}\cdots {\frac {1}{2}}W_{0}={\frac {(2p-1)!!}{(2p)!!}}\cdot {\frac {\pi }{2}}={\frac {(2p)!}{2^{2p}(p!)^{2}}}\cdot {\frac {\pi }{2}},}$
• ${\displaystyle W_{2p+1}={\frac {2p}{2p+1}}\cdot {\frac {2p-2}{2p-1}}\cdots {\frac {2}{3}}W_{1}={\frac {(2p)!!}{(2p+1)!!}}={\frac {2^{2p}(p!)^{2}}{(2p+1)!}}.}$

## Another relation to evaluate the Wallis' integrals

Wallis's integrals can be evaluated by using Euler integrals:

1. Euler integral of the first kind: the Beta function:
${\displaystyle \mathrm {B} (x,y)=\int _{0}^{1}t^{x-1}(1-t)^{y-1}\,dt={\frac {\Gamma (x)\Gamma (y)}{\Gamma (x+y)}}}$ for Re(x), Re(y) > 0
2. Euler integral of the second kind: the Gamma function:
${\displaystyle \Gamma (z)=\int _{0}^{\infty }t^{z-1}e^{-t}\,dt}$ for Re(z) > 0.

If we make the following substitution inside the Beta function: ${\displaystyle \quad \left\{{\begin{matrix}t=\sin ^{2}u\\1-t=\cos ^{2}u\\dt=2\sin u\cos udu\end{matrix}}\right.}$
we obtain:

${\displaystyle \mathrm {B} (a,b)=2\int _{0}^{\frac {\pi }{2}}\sin ^{2a-1}u\cos ^{2b-1}u\,du,}$

so this gives us the following relation to evaluate the Wallis integrals:

${\displaystyle W_{n}={\frac {1}{2}}\mathrm {B} \left({\frac {n+1}{2}},{\frac {1}{2}}\right)={\frac {\Gamma \left({\tfrac {n+1}{2}}\right)\Gamma \left({\tfrac {1}{2}}\right)}{2\,\Gamma \left({\tfrac {n}{2}}+1\right)}}.}$

So, for odd ${\displaystyle n}$, writing ${\displaystyle n=2p+1}$, we have:

${\displaystyle W_{2p+1}={\frac {\Gamma \left(p+1\right)\Gamma \left({\frac {1}{2}}\right)}{2\,\Gamma \left(p+1+{\frac {1}{2}}\right)}}={\frac {p!\Gamma \left({\frac {1}{2}}\right)}{(2p+1)\,\Gamma \left(p+{\frac {1}{2}}\right)}}={\frac {2^{p}\;p!}{(2p+1)!!}}={\frac {2^{2\,p}\;(p!)^{2}}{(2p+1)!}}}$

whereas for even ${\displaystyle n}$, writing ${\displaystyle n=2p}$ and knowing that ${\displaystyle \Gamma ({\tfrac {1}{2}})={\sqrt {\pi }}}$, we get :

${\displaystyle W_{2p}={\frac {\Gamma \left(p+{\frac {1}{2}}\right)\Gamma \left({\frac {1}{2}}\right)}{2\,\Gamma \left(p+1\right)}}={\frac {(2p-1)!!\;\pi }{2^{p+1}\;p!}}={\frac {(2p)!}{2^{2\,p}\;(p!)^{2}}}\cdot {\frac {\pi }{2}}.}$

## Equivalence

• From the recurrence formula above ${\displaystyle \mathbf {(2)} }$, we can deduce that
${\displaystyle \ W_{n+1}\sim W_{n}}$ (equivalence of two sequences).
Indeed, for all ${\displaystyle n\in \,\mathbb {N} }$ :
${\displaystyle \ W_{n+2}\leqslant W_{n+1}\leqslant W_{n}}$ (since the sequence is decreasing)
${\displaystyle {\frac {W_{n+2}}{W_{n}}}\leqslant {\frac {W_{n+1}}{W_{n}}}\leqslant 1}$ (since ${\displaystyle \ W_{n}>0}$)
${\displaystyle {\frac {n+1}{n+2}}\leqslant {\frac {W_{n+1}}{W_{n}}}\leqslant 1}$ (by equation ${\displaystyle \mathbf {(2)} }$).
By the sandwich theorem, we conclude that ${\displaystyle {\frac {W_{n+1}}{W_{n}}}\to 1}$, and hence ${\displaystyle \ W_{n+1}\sim W_{n}}$.
• By examining ${\displaystyle W_{n}W_{n+1}}$, one obtains the following equivalence:
${\displaystyle W_{n}\sim {\sqrt {\frac {\pi }{2\,n}}}\quad }$ ( and consequently ${\displaystyle \quad \lim _{n\rightarrow \infty }{\sqrt {n}}\,W_{n}={\sqrt {\pi /2}}\quad }$ ).
Proof

For all ${\displaystyle n\in \,\mathbb {N} }$, let ${\displaystyle u_{n}=(n+1)\,W_{n}\,W_{n+1}}$.

It turns out that, ${\displaystyle \forall n\in \mathbb {N} ,\,u_{n+1}=u_{n}}$ because of equation ${\displaystyle \mathbf {(2)} }$. In other words ${\displaystyle \ (u_{n})}$ is a constant.

It follows that for all ${\displaystyle n\in \,\mathbb {N} }$, ${\displaystyle u_{n}=u_{0}=W_{0}\,W_{1}={\frac {\pi }{2}}}$.

Now, since ${\displaystyle \ n+1\sim n}$ and ${\displaystyle \ W_{n+1}\sim W_{n}}$, we have, by the product rules of equivalents, ${\displaystyle \ u_{n}\sim n\,W_{n}^{2}}$.

Thus, ${\displaystyle \ n\,W_{n}^{2}\sim {\frac {\pi }{2}}}$, from which the desired result follows (noting that ${\displaystyle \ W_{n}>0}$).

## Deducing Stirling's formula

Suppose that we have the following equivalence (known as Stirling's formula):

${\displaystyle n!\sim C{\sqrt {n}}\left({\frac {n}{e}}\right)^{n},}$

for some constant ${\displaystyle C}$ that we wish to determine. From above, we have

${\displaystyle W_{2p}\sim {\sqrt {\frac {\pi }{4p}}}={\frac {\sqrt {\pi }}{2{\sqrt {p}}}}}$ (equation (3))

Expanding ${\displaystyle W_{2p}}$ and using the formula above for the factorials, we get

{\displaystyle {\begin{aligned}W_{2p}&={\frac {(2p)!}{2^{2p}(p!)^{2}}}\cdot {\frac {\pi }{2}}\\&\sim {\frac {C\left({\frac {2p}{e}}\right)^{2p}{\sqrt {2p}}}{2^{2p}C^{2}\left({\frac {p}{e}}\right)^{2p}\left({\sqrt {p}}\right)^{2}}}\cdot {\frac {\pi }{2}}\\&={\frac {\pi }{C{\sqrt {2p}}}}.{\text{ (equation (4))}}\end{aligned}}}

From (3) and (4), we obtain by transitivity:

${\displaystyle {\frac {\pi }{C{\sqrt {2p}}}}\sim {\frac {\sqrt {\pi }}{2{\sqrt {p}}}}.}$

Solving for ${\displaystyle C}$ gives ${\displaystyle C={\sqrt {2\pi }}.}$ In other words,

${\displaystyle n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}.}$

## Evaluating the Gaussian Integral

The Gaussian integral can be evaluated through the use of Wallis' integrals.

We first prove the following inequalities:

• ${\displaystyle \forall n\in \mathbb {N} ^{*}\quad \forall u\in \mathbb {R} _{+}\quad u\leqslant n\quad \Rightarrow \quad (1-u/n)^{n}\leqslant e^{-u}}$
• ${\displaystyle \forall n\in \mathbb {N} ^{*}\quad \forall u\in \mathbb {R} _{+}\qquad e^{-u}\leqslant (1+u/n)^{-n}}$

In fact, letting ${\displaystyle \quad u/n=t}$, the first inequality (in which ${\displaystyle t\in [0,1]}$) is equivalent to ${\displaystyle 1-t\leqslant e^{-t}}$; whereas the second inequality reduces to ${\displaystyle e^{-t}\leqslant (1+t)^{-1}}$, which becomes ${\displaystyle e^{t}\geqslant 1+t}$. These 2 latter inequalities follow from the convexity of the exponential function (or from an analysis of the function ${\displaystyle t\mapsto e^{t}-1-t}$).

Letting ${\displaystyle u=x^{2}}$ and making use of the basic properties of improper integrals (the convergence of the integrals is obvious), we obtain the inequalities:

${\displaystyle \int _{0}^{\sqrt {n}}(1-x^{2}/n)^{n}dx\leqslant \int _{0}^{\sqrt {n}}e^{-x^{2}}dx\leqslant \int _{0}^{+\infty }e^{-x^{2}}dx\leqslant \int _{0}^{+\infty }(1+x^{2}/n)^{-n}dx}$ for use with the sandwich theorem (as ${\displaystyle n\to \infty }$).

The first and last integrals can be evaluated easily using Wallis' integrals. For the first one, let ${\displaystyle x={\sqrt {n}}\,\sin \,t}$ (t varying from 0 to ${\displaystyle \pi /2}$). Then, the integral becomes ${\displaystyle {\sqrt {n}}\,W_{2n+1}}$. For the last integral, let ${\displaystyle x={\sqrt {n}}\,\tan \,t}$ (t varying from ${\displaystyle 0}$ to ${\displaystyle \pi /2}$). Then, it becomes ${\displaystyle {\sqrt {n}}\,W_{2n-2}}$.

As we have shown before, ${\displaystyle \lim _{n\rightarrow +\infty }{\sqrt {n}}\;W_{n}={\sqrt {\pi /2}}}$. So, it follows that ${\displaystyle \int _{0}^{+\infty }e^{-x^{2}}dx={\sqrt {\pi }}/2}$.

Remark: There are other methods of evaluating the Gaussian integral. Some of them are more direct.

## Note

The same properties lead to Wallis product, which expresses ${\displaystyle {\frac {\pi }{2}}\,}$ (see ${\displaystyle \pi }$) in the form of an infinite product.